Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int n, m, q;
struct node {
  int to, nxt;
} edge[N * 2];
int head[N], cnt, deep[N];
struct lca {
  int to, people[12], num;
  void init() { num = 0; }
  void add(int x) {
    if (num >= 10) return;
    people[num++] = x;
  }
  lca operator+(const lca &b) {
    lca tt;
    tt.init();
    int &t = tt.num;
    int i = 0, j = 0;
    while (t < 10 && i < num && j < b.num) {
      if (people[i] <= b.people[j])
        tt.people[t++] = people[i++];
      else
        tt.people[t++] = b.people[j++];
    }
    while (t < 10 && i < num) tt.people[t++] = people[i++];
    while (t < 10 && j < b.num) tt.people[t++] = b.people[j++];
    if (deep[to] < deep[b.to])
      tt.to = to;
    else
      tt.to = b.to;
    return tt;
  }
} par[N][20], self[N];
void init() {
  for (int i = 1; i <= n; i++)
    head[i] = deep[i] = -1, self[i].init(), self[i].to = i;
  cnt = 0;
}
void addEdge(int from, int to) {
  edge[cnt].to = to;
  edge[cnt].nxt = head[from];
  head[from] = cnt++;
}
void dfs(int u, int fa, int c) {
  deep[u] = c;
  par[u][0].to = fa;
  int v;
  for (int i = head[u]; i != -1; i = edge[i].nxt) {
    v = edge[i].to;
    if (v == fa) continue;
    dfs(v, u, c + 1);
  }
}
void calc() {
  int k = 0;
  for (int i = 0; i < 20; i++)
    if (n >> i & 1) k = i;
  for (int i = 2; i <= n; i++) par[i][0] = self[par[i][0].to];
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= n; j++) {
      int to = par[j][i - 1].to;
      if (deep[j] >= 1 << i) par[j][i] = par[j][i - 1] + par[to][i - 1];
    }
  }
}
void solve(int u, int v, int a) {
  if (deep[u] < deep[v]) swap(u, v);
  lca l, r;
  l = self[u], r = self[v];
  int dis = deep[u] - deep[v];
  for (int i = 0; 1 << i <= dis; i++)
    if (dis >> i & 1) l = l + par[u][i], u = par[u][i].to;
  if (u == v) {
    int k = min(l.num, a);
    printf("%d", k);
    for (int i = 0; i < k; i++) printf(" %d", l.people[i]);
    printf("\n");
    return;
  }
  for (int i = 17; i >= 0; i--) {
    if (deep[u] < 1 << i) continue;
    if (par[u][i].to == par[v][i].to) continue;
    l = l + par[u][i];
    r = r + par[v][i];
    u = par[u][i].to;
    v = par[v][i].to;
  }
  l = l + self[par[u][0].to] + r;
  int k = min(l.num, a);
  printf("%d", k);
  for (int i = 0; i < k; i++) printf(" %d", l.people[i]);
  printf("\n");
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  init();
  int a, b;
  for (int i = 1; i < n; i++)
    scanf("%d%d", &a, &b), addEdge(a, b), addEdge(b, a);
  for (int i = 1; i <= m; i++) scanf("%d", &a), self[a].add(i);
  dfs(1, 0, 0);
  calc();
  int u, v;
  while (q--) {
    scanf("%d%d%d", &u, &v, &a);
    solve(u, v, a);
  }
  return 0;
}